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Home » If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1Å, and (ii) R = 10 Å.
If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when
(i) R = 0.1Å,
and (ii) R = 10 Å.
Atoms | CBSE | Class 12 | ncert exemplar | Physics
If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when
(i) R = 0.1Å,
and (ii) R = 10 Å.

(i) Consider the nucleus of a H atom as a point charge electron circling about it at a speed of v and a radius of r_A The Coulombian force acts as a centrifugal force, causing the nucleus to circle.

\therefore \frac{\mathrm{m}_{\mathrm{e}} \mathrm{v}}{\mathrm{r}_{\mathrm{A}}}=\frac{-\mathrm{Ke}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}} \ldots(\mathrm{I})

Here \mathrm{K}=\frac{1}{4 \pi \varepsilon_{0}}

(-) sign shows the force of attraction.

By Bohr’s postulate, we have angular momentum as \frac{\mathrm{nh}}{2 \pi}

\mathrm{mvr}_{\mathrm{A}}=\frac{\mathrm{nh}}{2 \pi}

v=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}_{\mathrm{A}}}

\frac{m n^{2} h^{2}}{4 \pi^{2} \mathrm{~m}^{2} \mathrm{r}_{\mathrm{A}}^{2} \mathrm{r}_{\mathrm{A}}}=\frac{\mathrm{Ke}^{2}}{\mathrm{r}_{\mathrm{A}}^{2}}[ From I ]

\mathrm{r}_{\mathrm{A}}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2} \mathrm{mK} \mathrm{e}^{2}}

For ground state \mathrm{n}-1, we have,

r_{A}=\frac{h^{2}}{4 \pi^{2} \mathrm{mKe}^{2}}

=\frac{6.63 \times 10^{-34} 6.63 \times 10^{-34}}{(2 \times 3.14)^{2} 9.1 \times 10^{-31} \times 9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}

\frac{6.63 \times 6.63 \times 10^{-68+38+31-9}}{9.19 \times 1.6 \times 1.6 \times 4 \times 3.14 \times 4 \times 3.14}

=\mathrm{r}_{\mathrm{A}}=0.53 \mathrm{~A}^{\circ}

\begin{array}{l} \text { P.E. }=\frac{-\mathrm{Ke}^{2}}{\mathrm{r}_{\mathrm{A}}}=\frac{-9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{0.53 \times 10^{-10}} \mathrm{~J} \\ =\frac{-9 \times 1.6 \times 1.6 \times 10^{-19} \times 10^{-19} \times 10^{9}}{0.53 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{~J} \end{array}

\begin{array}{l} \text { P.E. }=\frac{14.4}{0.53}-27.17-27.2 \mathrm{eV} \\ \mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2} \end{array}

=\frac{1}{2} \frac{\mathrm{m} \cdot \mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2} \mathrm{~m}^{2} \mathrm{r}^{2}}=\frac{1}{2} \frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{mr}^{2}}(\mathrm{n}=1 for ground state)

=\frac{1}{2} \frac{6.62 \times 10^{-34} \times 6.62 \times 10^{-34}}{4 \times 3.14 \times 3.14 \times 9 \times 10^{-31} \times 0.53 \times 10^{-10} \times 0.53 \times 10^{-10}} \mathrm{~J}

=\frac{6.62 \times 6.62 \times 10^{-58+31+10+10}}{4 \times 3.14 \times 3.14 \times 18 \times 0.53 \times 0.53 \times 1.6 \times 10^{-19}} \mathrm{eV}

=\frac{6.62 \times 6.62 \times 10^{-58+51+19}}{4 \times 3.14 \times 3.14 \times 18 \times 0.53 \times 0.53}

\mathrm{K} . \mathrm{E}=0.1373 \times 10^{2} \mathrm{eV}=13.7 \mathrm{eV}

\mathrm{P} . \mathrm{E} .=27.2 \mathrm{eV}

(ii) In the case of a spherical nucleus with a radius of R, the electron moves charge within the nucleus R>>r_b, and then the electron moves inside the nucleus. Then ( r_b is the radius of the rotating electron’s new Bohr’s orbit)

Charge =\frac{\mathrm{e} \cdot\left(\frac{4}{3} \pi \mathrm{r}_{\mathrm{b}}^{3}\right)}{\frac{4}{4} \pi \mathrm{R}^{3}}

\mathrm{e}^{\prime}=\mathrm{q}_{2}=\frac{\mathrm{er}_{3}^{3}}{\mathrm{R}^{3}}

\mathrm{q}_{1}=\mathrm{e}

\frac{\mathrm{mv}^{2}}{\mathrm{r}_{3}}=\frac{\mathrm{Kee}^{\prime}}{\mathrm{r}_{3}^{2}}(\mathrm{By} Coulomb’s law )

\mathrm{m} v \mathrm{r}_{3}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \frac{\mathrm{nh}}{2 \pi \mathrm{mr}_{3}} (By Bohr’s postulate)

\begin{array}{l} \therefore \frac{m}{r_{b}} \frac{n^{2} h^{2}}{4 \pi^{2} m^{2} r_{3}^{2}}=\frac{k e e^{\prime}}{r_{b}^{2}} \\ r_{b}=\frac{n^{2} h^{2}}{4 \pi^{2} m k e e^{\prime}} \end{array}

Now for ground state of \mathrm{H}, \mathrm{m}=1 and \mathrm{e}^{\prime}=\frac{\mathrm{er}_{\mathrm{A}}^{3}}{\mathrm{R}^{3}}, then we have,

\therefore \mathrm{r}_{\mathrm{b}}=\frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{mK} \cdot \mathrm{e} \cdot \mathrm{e} \cdot \frac{\mathrm{r}_{\mathrm{b}}^{3}}{\mathrm{R}^{3}}}=\left[\frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{mK} \mathrm{e}^{2}}\right] \times \frac{\mathrm{R}^{3}}{\mathrm{r}_{\mathrm{b}}^{3}}=\mathrm{r}_{\mathrm{A}} \times \frac{\mathrm{R}^{3}}{\mathrm{r}_{\mathrm{b}}^{3}}

\begin{array}{l} \left[\because \mathrm{r}_{\mathrm{A}}=\frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{mK} \mathrm{e}^{3}}=0.53 \mathrm{~A}^{\circ} \text { calclated in } \operatorname{part}(\mathrm{i})\right]\\ \mathrm{r}_{\mathrm{b}}=\mathrm{r}_{\mathrm{A}}\left[\frac{\mathrm{R}}{\mathrm{r}_{\mathrm{b}}}\right]^{3}\\ \mathrm{r}_{\mathrm{b}}^{4}=\mathrm{r}_{\mathrm{A}} \mathrm{R}^{3}=0.53 \mathrm{~A}^{\circ} \times\left(10 \mathrm{~A}^{\circ}\right)^{3}\\ \mathrm{r}_{\mathrm{b}}^{\mathrm{A}}=0.53 \times 1000\left(\mathrm{~A}^{\circ}\right)^{4}\left(\because \mathrm{r}_{\mathrm{A}}=0.53 \mathrm{~A}^{\circ}\right)\\ \mathrm{r}_{\mathrm{b}}=[530] \frac{1}{4} \mathrm{~A}^{\circ}<\mathrm{RA}^{\circ}\\ \text { K.E. }=\frac{1}{2} m v^{2}=\frac{m}{2} \cdot \frac{h^{2}}{4 \pi^{2} \mathrm{~m}^{2} \mathrm{r}_{\mathrm{b}}^{2}}=\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{mr}_{\mathrm{b}}^{2}}\left[\because \mathrm{v}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2} \mathrm{~m}^{2} \mathrm{r}_{\mathrm{b}}^{2}}\right]\\ =\frac{6.62 \times 6.62 \times 10^{-34} \times 10^{-34}}{8 \times 3.14 \times 3.14 \times 9.1 \times 10^{-31} \times 4.8 \times 4.8 \times 10^{-20} \times 1.6 \times 10^{-19}} \mathrm{eV}\\ =\frac{6.62 \times 6.62 \times 10^{-68+31+20+19}}{8 \times 3.14 \times 3.14 \times 9.1 \times 4.8 \times 4.8 \times 1.6} \mathrm{eV}\\ =\frac{43.8244}{26460.2} 10^{-58+70}-0.001656 \times 10^{2} \mathrm{eV}\\ \mathrm{K} . \mathrm{E}=0.167 \mathrm{eV}\\ \mathrm{P} . \mathrm{E} .=\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}} \frac{\left(\mathrm{r}_{\mathrm{b}}^{2}-3 \mathrm{R}^{2}\right)}{\mathrm{R}^{3}}\left[\mathrm{P} \cdot \mathrm{E}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}}\right]\\ \mathrm{P} . \mathrm{E} .=\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} \mathrm{r}_{\mathrm{A}}}\right] \frac{\mathrm{r}_{\mathrm{A}}\left(\mathrm{r}_{\mathrm{b}}^{2}-3 \mathrm{R}^{2}\right)}{\mathrm{R}^{3}}\left[\text { multiplying by } \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{A}}}\right] \end{array}

From part (i), we have,

\mathrm{P} . \mathrm{E} .=\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}_{\mathrm{A}}}=27.2 \mathrm{eV}

\therefore \mathrm{P} \cdot \mathrm{E}=27.2\left[\frac{0.53(\sqrt{530}-300)}{1000}\right]\left[\because \mathrm{r}_{\mathrm{b}}=(530)^{\frac{1}{4}} \mathrm{~A}^{\circ} \text { and } \mathrm{R}=10 \mathrm{~A}^{\circ}\right]

=\frac{27.2 \mathrm{eV} \times 0.53(23.02-300) \mathrm{A}^{3}}{1000 \mathrm{~A}^{3}}=27.2 \times \frac{0.53(-276.9)}{1000} \mathrm{eV}

\mathrm{P} . \mathrm{E} .=\frac{3992.9}{1000}=-3.99 \mathrm{eV}

\mathrm{K} . \mathrm{E} .=0.167 \mathrm{eV}

i

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